Independent solution

How to solve this Bayes’ Theorem question

Setup

Setup

Within the stated three-model-year restriction, multiply each year's prior share by its conditional accident probability.

w14=0.16(0.05)=0.0080w_{14}=0.16(0.05)=0.0080
w13=0.18(0.02)=0.0036w_{13}=0.18(0.02)=0.0036
w12=0.20(0.03)=0.0060w_{12}=0.20(0.03)=0.0060

Model

Model

Condition on both an accident and membership in the three eligible years by normalizing the target year's weight over the three weights.

Pr⁡(14∣A, Y∈{12,13,14})=w14w14+w13+w12\Pr(14\mid A,\,Y\in\{12,13,14\})=\frac{w_{14}}{w_{14}+w_{13}+w_{12}}

Compute

Compute

The target weight is 0.0080 and the restricted accident weights total 0.0176, so the posterior is 0.0080/0.0176=0.454545.

0.00800.0176=0.454545\frac{0.0080}{0.0176}=0.454545

Answer

Answer

The requested model-year probability rounds to 0.45, corresponding to choice D.

0.45(D)\boxed{0.45\quad\text{(D)}}